JEE Main 2023PhysicsAtomic PhysicsMediumNumerical

JEE Main 2023 — Atomic Physics Question with Solution

JEE Main 2023 (01 Feb Shift 2)

Question

Nucleus a having Z=17 and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon. Another nucleus B of Z=12 has total 26 nucleons and 1.8 MeV binding energy per nucleons. The difference of binding energy of B and A will be ______ MeV.

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

In A, Z=17 and have equal numbers of protons and neutrons i.e. nucleons=2×17=34

So, for A, 

Total binding energy BEA=1.2×34=40.8 MeV

Another nucleus B mass number =26

So, total binding energy of  B is BEB=1.8×26 MeV=46.8 MeV

Difference of BE=BEB-BEA=6 MeV

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About this question

This is a previous-year question from JEE Main 2023, covering the Atomic Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.